Sets, Relations and Functions
This chapter looks like definitions and is actually the grammar of the whole paper. Every later topic — inverse trigonometry, limits, integration, probability — depends on knowing precisely what a function's domain is and whether it is one-one or onto, because those two properties decide whether an inverse exists at all. The single most productive habit to build here is to write down the domain before doing anything else with a function. Most lost marks in this chapter are not conceptual failures but omissions: a square root whose argument can go negative, a denominator that can vanish, a logarithm of something not strictly positive.
Relations and Equivalence
| Property | Condition | Example on integers |
|---|---|---|
| Reflexive | (a, a) ∈ R for every a | "a ≤ b" is reflexive; "a < b" is not |
| Symmetric | (a, b) ∈ R implies (b, a) ∈ R | "a = b" is symmetric; "a ≤ b" is not |
| Transitive | (a, b) and (b, c) in R implies (a, c) in R | "a divides b" is transitive |
| Equivalence | All three at once | "a ≡ b (mod 5)" — partitions ℤ into 5 classes |
⚠️ The empty relation is symmetric and transitive but not reflexive
This is a favourite because it exposes whether you read the definitions as conditions or as slogans. Symmetry and transitivity are conditional statements — "if (a,b) is in R then …" — and a conditional with a false premise is automatically true, so a relation with no ordered pairs at all satisfies both vacuously. Reflexivity is not conditional: it demands that (a,a) be present for every a, so the empty relation fails it on any non-empty set. The mirror case is also examined: on a set with one element, the empty relation is still not reflexive, while the relation {(a,a)} is an equivalence relation. Read each definition as the logical statement it is, and these questions become mechanical.Functions, Domain and Invertibility
| Expression | Requirement |
|---|---|
| √f(x) | f(x) ≥ 0 |
| 1/f(x) | f(x) ≠ 0 |
| log f(x) | f(x) > 0 — strictly, not ≥ 0 |
| log_g(x) f(x) | f > 0, g > 0 and g ≠ 1 |
| sin⁻¹x or cos⁻¹x | −1 ≤ x ≤ 1 |
| tan x | x ≠ (2n + 1)π/2 |
🎯 Why a function needs to be both one-one and onto to have an inverse
An inverse must send every element of the codomain back to exactly one element of the domain. If the function is not one-one, two inputs share an output and the inverse would not know which to return — that is why f(x) = x² has no inverse on ℝ but does on [0, ∞). If it is not onto, some codomain element has no pre-image and the inverse would be undefined there — which is why f(x) = eˣ is not invertible as a map to ℝ but is invertible as a map to (0, ∞). So restricting the domain fixes one-one-ness and restricting the codomain fixes ontoness, and this is exactly why inverse trigonometric functions come with principal-value ranges rather than being defined outright. A question that asks you to "make f invertible" is asking which of the two restrictions is needed.Counting maps between finite sets is a standard question type. From a set of m elements to one of n elements there are nᵐ functions in all, n(n−1)…(n−m+1) one-one functions (zero if m > n), and for onto functions the count comes from inclusion–exclusion. Two easy checks save time: if m > n no function can be one-one, and if m < n none can be onto — so half the options can often be eliminated before any arithmetic.
🧠 Test even and odd by substituting −x, and use it to kill integrals
A function is even if f(−x) = f(x) and odd if f(−x) = −f(x); most functions are neither. The payoff comes much later but is worth setting up now: the integral of an odd function over a symmetric interval [−a, a] is exactly zero, so recognising oddness converts a hard definite integral into a one-line answer. Note also that sin x and tan x are odd while cos x is even, that the product of two odd functions is even, and that the only function that is both even and odd is f(x) = 0. Spotting symmetry first is one of the highest-yield habits in the whole paper.Key takeaways
- A set of n elements has 2ⁿ subsets; there are 2^(mn) relations from an m-set to an n-set.
- Inclusion–exclusion adds odd-sized intersections and subtracts even-sized ones.
- "Exactly two" is not "at least two" — subtract three times the triple intersection.
- The empty relation is symmetric and transitive vacuously, but never reflexive on a non-empty set.
- Always write the domain first: √ needs ≥ 0, a denominator needs ≠ 0, log needs strictly > 0.
- An inverse exists only if f is one-one and onto — restrict the domain, or the codomain, or both.
- From an m-set to an n-set: nᵐ functions in all, and no one-one map exists if m > n.
- The integral of an odd function over [−a, a] is zero — spot symmetry before integrating.
Practice questions (8)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
The number of relations from a set with 3 elements to a set with 2 elements is:
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Answer: C — 64
A relation is any subset of the Cartesian product, which here has 3 × 2 = 6 ordered pairs, so the number of subsets is 2⁶ = 64. The answer 8 is 2³, which counts subsets of the first set alone, and 6 is the size of the product rather than the number of its subsets — both are the kind of near-miss that comes from stopping one step early.The empty relation on a non-empty set A is:
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Answer: B — Symmetric and transitive but not reflexive
Symmetry and transitivity are conditional statements, and with no ordered pairs the premises are never satisfied, so both hold vacuously. Reflexivity is unconditional — it requires (a, a) for every a in A — and the empty relation supplies none of them, so it fails. Hence the relation is not an equivalence relation despite satisfying two of the three requirements.The domain of f(x) = log(x − 2) + √(5 − x) is:
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Answer: B — (2, 5]
The logarithm needs x − 2 > 0 strictly, so x > 2 and the endpoint 2 is excluded; the square root needs 5 − x ≥ 0, so x ≤ 5 and the endpoint 5 is included. Intersecting gives (2, 5]. The trap is treating the logarithm like the square root and writing [2, 5] — log requires a strictly positive argument, since log 0 is undefined.f : ℝ → ℝ given by f(x) = x² is:
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Answer: C — Neither one-one nor onto
It is not one-one because f(2) = f(−2) = 4, and not onto because no real number squares to a negative value, so the whole negative half of the codomain is unreached. Restricting the domain to [0, ∞) makes it one-one and restricting the codomain to [0, ∞) makes it onto — do both and the function becomes invertible, which is exactly how √x is defined.In a group of 100 people, 65 read A, 45 read B and 20 read both. How many read neither?
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Answer: A — 10
By inclusion–exclusion n(A ∪ B) = 65 + 45 − 20 = 90, so 100 − 90 = 10 read neither. Adding 65 and 45 without subtracting the overlap gives 110, more than the group itself, which is an immediate signal that the intersection must be removed — a useful sanity check whenever the subtotals exceed the total.The number of one-one functions from a set with 4 elements to a set with 3 elements is:
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Answer: C — 0
A one-one function needs a distinct image for each of the four inputs, but the codomain offers only three values, so by the pigeonhole principle two inputs must collide and no such function exists. The answer 81 is 3⁴, the total number of functions, which is a different count entirely — checking whether m > n before calculating anything settles questions of this type instantly.The relation "a ≡ b (mod 3)" on ℤ is:
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Answer: B — An equivalence relation partitioning ℤ into 3 classes
Every integer is congruent to itself so the relation is reflexive; if a − b is divisible by 3 then so is b − a, giving symmetry; and if 3 divides both a − b and b − c it divides their sum a − c, giving transitivity. Being an equivalence relation, it partitions the integers into the three residue classes of remainder 0, 1 and 2 — and every equivalence relation partitions its set in just this way.Which function is odd?
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Answer: B — f(x) = x³ + x
Substituting −x gives (−x)³ + (−x) = −(x³ + x) = −f(x), which is the definition of odd. The other three all satisfy f(−x) = f(x) and are therefore even. Recognising oddness is worth the check because the integral of an odd function over a symmetric interval such as [−a, a] vanishes, turning a laborious integration into a single step.